A227870 Numbers with equal number of even and odd digits.

10, 12, 14, 16, 18, 21, 23, 25, 27, 29, 30, 32, 34, 36, 38, 41, 43, 45, 47, 49, 50, 52, 54, 56, 58, 61, 63, 65, 67, 69, 70, 72, 74, 76, 78, 81, 83, 85, 87, 89, 90, 92, 94, 96, 98, 1001, 1003, 1005, 1007, 1009, 1010, 1012, 1014, 1016, 1018, 1021, 1023, 1025

Offset

1,1

Comments

Numbers with an odd digit length cannot be in this sequence. - Alonso del Arte, Nov 02 2013

Links

Amiram Eldar, Table of n, a(n) for n = 1..10000

Example

1009 has 2 even digits (00) and 2 odd digits (19) and so is in the sequence.

Mathematica

Select[Range[1025], (d = Differences[Tally[Mod[IntegerDigits[#], 2]]]) != {} && d[[1, 2]] == 0 &] (* Amiram Eldar, Oct 01 2020 *)
eneodQ[n_]:=With[{id=IntegerDigits[n]}, Count[id, _?(OddQ[#]&)]==Count[id, _?(EvenQ[#]&)]]; Select[Range[1100], eneodQ] (* Harvey P. Dale, Jul 19 2024 *)

Prog

(JavaScript)
for (i = 1; i < 5000; i++) {
s = i.toString();
odds = 0; evens = 0;
for (j = 0; j < s.length; j++) if (s.charAt(j)%2 == 0) evens++; else odds++;
if (odds == evens) document.write(i + ", ");
}
(PARI) isok(m) = my(d=digits(m)); #select(x->(x%2), d) == #select(x->!(x%2), d); \\ Michel Marcus, Oct 01 2020
(Python)
def ok(i):
  stri = str(i)
  se = sum(1 for d in stri if d in "02468")
  so = sum(1 for d in stri if d in "13579")
  return se == so
def aupto(nn):
  alst, an = [None], 0
  for n in range(1, nn+1):
    while len(alst) < nn+1:
      if ok(an): alst.append(an)
      an += 1
  return alst[1:] # use alst[n] for a(n)
print(aupto(58))  # Michael S. Branicky, Dec 14 2020

Crossrefs

Subsequence of A001637.

Cf. A030141, A031443.

Sequence in context: A180157 A309539 A273492 * A007958 A179083 A092132

Adjacent sequences: A227867 A227868 A227869 * A227871 A227872 A227873

Keyword

nonn,base,easy,changed

Author

Jon Perry, Nov 02 2013

Status

approved